| Exam Board | Edexcel |
|---|---|
| Module | FP2 AS (Further Pure 2 AS) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Number Theory |
| Type | Modular arithmetic properties |
| Difficulty | Standard +0.3 This is a straightforward modular arithmetic question requiring standard techniques: (i) uses basic modular reduction (214 ≡ 6 mod 8, then 6^6 ≡ 0 mod 8), and (ii) applies divisibility rules for 11 and 3. While it's Further Maths content, these are routine applications of well-known divisibility tests with minimal problem-solving required, making it slightly easier than average overall. |
| Spec | 8.02b Divisibility tests: standard tests for 2, 3, 4, 5, 8, 9, 118.02e Finite (modular) arithmetic: integers modulo n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(214 \equiv 6 \pmod{8}\) | B1 | |
| \((214)^2 \equiv 6^2 \pmod{8} \Rightarrow (214)^2 \equiv 4 \pmod{8}\); or \((214)^6 \equiv 6^6 \pmod{8} \equiv (6^3)^2 \pmod{8} \equiv (0)^2 \pmod{8}\); or \((214)^3 \equiv 6^3 \pmod{8} \Rightarrow (214)^3 \equiv 0 \pmod{8}\); or \(214 \equiv -2 \pmod{8} \Rightarrow (214)^6 \equiv (-2)^6 \pmod{8} \equiv 64 \pmod{8}\) | M1 | Uses modulo arithmetic to find \((214)^6 \equiv a \pmod{8}\) |
| \((214)^6 \equiv 0 \pmod{8}\), therefore \(214^6\) is divisible by 8 | A1 | Must draw conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses \(a - 5 + b - 8 + a - b + 0 = 2a - 13 = 11n\) to find \(a\) | M1 | Uses division rule for 11 |
| \(a = 1\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1 + 5 + b + 8 + 1 + b + 0 = 2b + 15 = 3n\) to find \(b\) | M1 | Uses division rule for 3 |
| Any two of \(b = 0,\ 3,\ 6\) or \(9\) | A1 | |
| All four of \(b = 0,\ 3,\ 6\) and \(9\) | A1 |
# Question 5(i) — Modular Arithmetic
| Answer/Working | Mark | Guidance |
|---|---|---|
| $214 \equiv 6 \pmod{8}$ | B1 | |
| $(214)^2 \equiv 6^2 \pmod{8} \Rightarrow (214)^2 \equiv 4 \pmod{8}$; or $(214)^6 \equiv 6^6 \pmod{8} \equiv (6^3)^2 \pmod{8} \equiv (0)^2 \pmod{8}$; or $(214)^3 \equiv 6^3 \pmod{8} \Rightarrow (214)^3 \equiv 0 \pmod{8}$; or $214 \equiv -2 \pmod{8} \Rightarrow (214)^6 \equiv (-2)^6 \pmod{8} \equiv 64 \pmod{8}$ | M1 | Uses modulo arithmetic to find $(214)^6 \equiv a \pmod{8}$ |
| $(214)^6 \equiv 0 \pmod{8}$, therefore $214^6$ is divisible by 8 | A1 | Must draw conclusion |
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# Question 5(ii)(a) — Divisibility by 11
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $a - 5 + b - 8 + a - b + 0 = 2a - 13 = 11n$ to find $a$ | M1 | Uses division rule for 11 |
| $a = 1$ | A1 | |
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# Question 5(ii)(b) — Divisibility by 3
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1 + 5 + b + 8 + 1 + b + 0 = 2b + 15 = 3n$ to find $b$ | M1 | Uses division rule for 3 |
| Any two of $b = 0,\ 3,\ 6$ or $9$ | A1 | |
| All four of $b = 0,\ 3,\ 6$ and $9$ | A1 | |\begin{enumerate}
\item (i) Making your reasoning clear and using modulo arithmetic, show that
\end{enumerate}
$$214 ^ { 6 } \text { is divisible by } 8$$
(ii) The following 7-digit number has four unknown digits
$$a 5 \square b \square a b 0$$
Given that the number is divisible by 11\\
(a) determine the value of the digit $a$.
Given that the number is also divisible by 3\\
(b) determine the possible values of the digit $b$.
\hfill \mbox{\textit{Edexcel FP2 AS 2023 Q5 [8]}}